Step 1

Given data

Error = 0.02

Confidence level = 0.95

Significance level \(\displaystyle=\alpha={1}-{0.95}={0.05}\)

\(\displaystyle{Z}_{{\frac{{0.05}}{{2}}}}={Z}_{{{0.025}}}=\pm{1.96}\)

(From Excel = NORM.S.INV(0.025))

sample proportion(p)=0.30

a)

Sample size is given by(n)

Margin of error formula is given by

\(\displaystyle{E}={Z}_{{\frac{\alpha}{{2}}}}\times\sqrt{{\frac{{{p}-{\left({1}-{p}\right)}}}{{n}}}}\)

Simplifying the above formula

\(\displaystyle{n}={p}{\left({1}-{p}\right)}\times{\left(\frac{{Z}_{{\frac{\alpha}{{2}}}}}{{E}}\right)}^{{2}}={0.3}\times{\left({1}-{0.3}\right)}\times{\left(\frac{{1.96}}{{0.02}}\right)}^{{2}}={2016.84}\approx{2017}\)

Step 2

b)

n=2017

No of smokers = 520

Point estimate is given by

\(\displaystyle\hat{{{p}}}=\frac{{520}}{{2017}}={0.258}\)

Step 3

c)

95% confidence interval for the proportion of smokers in the population is given by

Confidence level = 0.95

Significance level \(\displaystyle=\alpha={1}-{0.95}={0.05}\)

\(\displaystyle{Z}_{{\frac{{0.05}}{{2}}}}={Z}_{{0.025}}=\pm{1.96}\)

(From Excel = NORM.S.INV(0.025))

Confidence interval is given by

\(\displaystyle\hat{{{p}}}+{Z}_{{\frac{\alpha}{{2}}}}\sqrt{{\frac{{\hat{{{p}}}-{\left({1}-\hat{{{p}}}\right)}}}{{n}}}}={0.258}\pm{1.96}\sqrt{{\frac{{{0.258}{\left({1}-{0.258}\right)}}}{{2017}}}}={\left({0.239},{0.277}\right)}\)

Step 4

Result:

a) 2017

b) 0.258

c) (0.239,0.277)

Given data

Error = 0.02

Confidence level = 0.95

Significance level \(\displaystyle=\alpha={1}-{0.95}={0.05}\)

\(\displaystyle{Z}_{{\frac{{0.05}}{{2}}}}={Z}_{{{0.025}}}=\pm{1.96}\)

(From Excel = NORM.S.INV(0.025))

sample proportion(p)=0.30

a)

Sample size is given by(n)

Margin of error formula is given by

\(\displaystyle{E}={Z}_{{\frac{\alpha}{{2}}}}\times\sqrt{{\frac{{{p}-{\left({1}-{p}\right)}}}{{n}}}}\)

Simplifying the above formula

\(\displaystyle{n}={p}{\left({1}-{p}\right)}\times{\left(\frac{{Z}_{{\frac{\alpha}{{2}}}}}{{E}}\right)}^{{2}}={0.3}\times{\left({1}-{0.3}\right)}\times{\left(\frac{{1.96}}{{0.02}}\right)}^{{2}}={2016.84}\approx{2017}\)

Step 2

b)

n=2017

No of smokers = 520

Point estimate is given by

\(\displaystyle\hat{{{p}}}=\frac{{520}}{{2017}}={0.258}\)

Step 3

c)

95% confidence interval for the proportion of smokers in the population is given by

Confidence level = 0.95

Significance level \(\displaystyle=\alpha={1}-{0.95}={0.05}\)

\(\displaystyle{Z}_{{\frac{{0.05}}{{2}}}}={Z}_{{0.025}}=\pm{1.96}\)

(From Excel = NORM.S.INV(0.025))

Confidence interval is given by

\(\displaystyle\hat{{{p}}}+{Z}_{{\frac{\alpha}{{2}}}}\sqrt{{\frac{{\hat{{{p}}}-{\left({1}-\hat{{{p}}}\right)}}}{{n}}}}={0.258}\pm{1.96}\sqrt{{\frac{{{0.258}{\left({1}-{0.258}\right)}}}{{2017}}}}={\left({0.239},{0.277}\right)}\)

Step 4

Result:

a) 2017

b) 0.258

c) (0.239,0.277)